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ta The Expert TA Human-lik TA | Human-like https://us380k.theexpertta.com/Common/TakeTutorialAssignment.aspx Class Management Help Homework 7 Begin Date: 2/13/2016 12:00:00 AM -- Due Date: 2/27/2016 12:00:00 AM End Date: 2/27/2016 12:00:00 AM (10%) Problem 10: A rod of m: 0.35 kg and L = 0.55 m is on a ramp of two parallel rails with angle = 12° with respect to the horizontal. The current on the rod is I = 1.2 A, pointing into the screen as shown. A uniform magnetic field B- 0.55 T which points upward is applied in the region. Ignore the friction on the Randomized Variables ©theexpertta.com m=0.35 kg L=0.55 m =12° B=0.55 T I=1,2 A Assignment Status Can You Please Find The Answer To All The Parts. .. Chegg.com Click here for detailed view 10% Part (a) Express the magnitude of the magnetic force in terms of L, 1 and B 10% Part (b) Calculate the numerical value of the magnitude of the magnetic force in N 10% Part (c) Which direction is the magnetic force? 10% Part (d) Express the component of the magnetic force parallel to the ramp in terms of FB and Problem Status Completed Completed Completed Completed Completed Completed Completed Completed Completed Partial D 10% Part (e) Express the component of the weight parallel to the ramp in terms of m, g and . Take up the ramp to be positive 10% Part (f) Calculate the numerical value of the net force F parallel to the ramp, assuming the positive direction is up the ramp in N 4 D 10% Part (g) Express the acceleration along the ramp in terms of the F, and mt 2 10% Part (h) Calculate the numerical value of a in m/s 6 10% Part (i) If a is positive, which direction does the rod go? 10% Part (j) If a is negative, which direction does the rod go? Grade Summary Deductions Potential Down the ramp.OUp the ramp 9 10 0% 100% Submissions Attempts remaining: 1 % per attempt) detailed view Submit I give up! Hints: 0 for a 0% deduction. Hints remaining: 0 Feedback: 1% deduction per feedback

Solution

e) -mgsin(theta)
f) Force due to magnetic field,F = ILB in the right direction
Component parralel to the ramp = ILBcos(theta)
Net force || to ramp, Fn = ILBcos(theta)-mgsin(theta)
g) acceleration = Fn/m
h) a = [ILBcos(theta)-mgsin(theta)]/m = [1.2*0.55*0.55cos(12)-0.35*9.8*sin(12)]/0.35 = [0.355 - 0.713]/0.35 = -1.02 m/s^2
i) Up the ramp
j) down the ramp